University Physics 226N/231N Old Dominion University ...Aug 31, 2016 · Old Dominion University...

Prof. Satogata / Fall 2012 ODU University Physics 226N/231N 1

University Physics 226N/231N Old Dominion University

Kinematics in One Dimension

Dr. Todd Satogata (ODU/Jefferson Lab) [email protected]

http://www.toddsatogata.net/2016-ODU

Wednesday, August 31 2016 Happy Birthday to Richard Gere, Van Morrison, Frank Robinson,

Hermann Helmholtz, and David Politzer (Nobel Prize, 2004) Happy Eat Outside Day and National Trail Mix Day!

Please set your cell phones to “vibrate” or “silent” mode. Thanks!


What We Covered Last Time…

§  We will usually have one slide of “previous concepts review” at the start of every new class

§  For today, the most important concepts from Monday are: §  Nearly all physical quantities have units (m, s, kg, …)

•  Units can be treated algebraically, e.g. canceled in ratios

§  Pay attention to significant figures when writing results •  Carry all digits through calculations then round only final result •  There is a nice Wikipedia article on significance arithmetic

§  Example:

§  Draw pictures and tell stories when solving problems •  This keeps you organized and helps find a path to a solution •  This also helps “future you” remember what you did

§  We’ll return to vector algebra when we get into 2D motion •  We don’t need it much for 1D motion to be discussed today.

�x = v�t = (3.1 m/s)(3 s) = 9 m


Kinematics

§  Kinematics §  the branch of mechanics concerned with the motion of objects

without reference to the forces that cause the motion.

§  Primary concepts of kinematics §  Time

•  Things don’t move without time passing §  Position

•  Where have we been? Where are we going? (Are we there yet? Are we there yet? Are we there yet?)

§  Velocity •  How are changes in position and time related to each other?

§  Acceleration •  How are changes in velocity and time related to each other?

�x(t)

v(t)

�t

a(t)


Kinematics

§  Kinematics §  the branch of mechanics concerned with the motion of objects

without reference to the forces that cause the motion.

§  Primary concepts of kinematics §  Time

•  Things don’t move without time passing §  Position

•  Where have we been? Where are we going? (Are we there yet? Are we there yet? Are we there yet?)

§  Velocity •  How are changes in position and time related to each other?

§  Acceleration •  How are changes in velocity and time related to each other?

§  It turns out we don’t need to go beyond these – these are enough

�t

Vector functions of time �x(t) �y(t) �z(t)

vx

(t) vy

(t) vz

(t)

ax

(t) ay

(t) az

(t)


Motion in One Dimension: Position

§  Position : linear distance from a “zero” reference point §  To be general, we only discuss position differences or changes §  For example

§  I generally try to use instead of wherever I can •  is a positive or negative distance between two points

§  In these discussions we will be talking about falling objects a lot, so you will see both and

•  Try to get in the habit of drawing and labeling axes •  This helps make positive and negative signs obvious

x

�x = xfinal � xinitial = x(t2)� x(t1) = x2 � x1

�x x

�x

�x �y

��

��

��

��

��

��

��

��


§  Velocity: How far an object moves in a given time interval : §  Velocity, like position and displacement, is a vector with

magnitude and direction •  We use speed for just the magnitude in common language

§  Velocity, like position, is a function of time. Over a finite time period, we call this average velocity:

§  The logical extension from calc is a derivative §  Instantaneous velocity is the slope of position over a very

small time, as

We can figure this out for any time t, and it’s continuous like position

Motion in One Dimension: Velocity

�x

�t v = �x/�t

v(t) = �x(t)/�t

v(t) = �x(t)/�t (for very small�t)

�t

�x(t)=

x(t

2)�

x(t

1)


Velocity is a Slope of Position vs Time –  Velocity is the slope of the curve of x(t): how fast position is

changing with time. Note that it can be positive or negative!

§  That means that velocity is also a function and we can plot velocity v(t) like we plot position x(t) §  And then we can figure out how the velocity is changing (take slopes) too!


Acceleration is a Slope of Velocity vs Time

§  Acceleration is the rate of change of velocity. §  Exactly like velocity was the rate

of change of position!

§  Average velocity over a time interval ∆t is defined as the change in velocity divided by the time:

§  Instantaneous acceleration is the limit of the average acceleration as the time interval becomes arbitrarily short:

–  Acceleration is the slope of v(t)

a =�v

�t

a =�v

�t(for very small�t)


Ponderable Example

��

��

��

��

��

��

��

��

��

��

What are the velocity and acceleration of the red trajectory at t=0 s? Blue trajectory at t=1.5 s?


Position, Velocity, and Acceleration

§  Individual or absolute values of position, velocity, and acceleration are not related.

§  Instead, velocity depends on the rate of change of position. §  Acceleration depends on the rate of change of velocity.

•  These are all relative quantities, and not based on absolute position or position of the origin

•  This makes our description of this motion universal §  An object can be at position x = 0 and still be moving. §  An object can have zero velocity and still be accelerating.

§  At the peak of its trajectory, a juggling thud has §  Maximum vertical displacement from my hand §  Zero vertical velocity §  Constant (negative?) acceleration due to the force of gravity


Juggler Physics

y

�y > 0 vy ⇡ 0 ay < 0

�y > 0 vy > 0 ay < 0

�y = 0 vy ⇡ 0 ay ⇡ 0

vy

acceleration of gravity -9.8 m/s2


Also Juggler Physics!

y�y = 0 vy ⇡ 0 ay ⇡ 0

vy

�y < 0 vy < 0 ay > 0

�y < 0 vy ⇡ 0 ay > 0

acceleration of gravity +9.8 m/s2


Constant Acceleration

§  When acceleration is constant: position x, velocity v, acceleration a, and time t are related by where x0 and v0 are initial values at time t = 0 and x(t) and v(t) are the values at an arbitrary time t.

•  With constant acceleration –  Velocity is a linear

function of time –  Position is a quadratic

function of time �v(t) = at

�x(t) =1

2[v0 + v(t)]t

�x(t) = v0t+1

2at

2

v

2(t)� v

20 = 2a�x(t)


The Acceleration of Gravity

§  The acceleration of gravity at any point is (basically) the same for all objects, regardless of mass.

§  Near Earth’s surface, the magnitude of the acceleration is essentially constant at

§  Therefore the equations for constant acceleration apply: §  In a coordinate system with y axis

upward so , they read

This strobe photo of a falling ball shows increasing spacing resulting from the acceleration of gravity.

g = 9.8 m/s2 = 32 ft/s2

�v(t) = �gt

�y(t) =1

2[v0 + v(t)]t

�y(t) = v0t�1

2gt2

v2(t)� v20 = �2g�y(t)

~a = �gy


(Honors Ponderable)

§  Assume that the 13 ball in the photo is a “standard” 2.25 inch diameter billiard ball

§  How fast is the strobe flashing between images of the falling ball?

§  What is the ball’s approximate instantaneous velocity in the first image? In the last?

This strobe photo of a falling ball shows increasing spacing resulting from the acceleration of gravity.


Example: The Acceleration of Gravity

§  A ball is thrown straight up at 7.3 m/s, leaving your hand 1.5 m above the ground. Find its maximum height and when it hits the floor. §  At the maximum height the ball

is instantaneously at rest (even though it’s still accelerating). Solving the last equation with v = 0 gives the maximum height:

( )20 0

2 20

0 2

0 2or

(7.3 m/s)1.5 m 4.2 m2 (2)(9.8 m/s )

v g y y

vy yg

= − −

= + = + =

•  Setting y = 0 in the third equation gives a quadratic in time; the result is the two values for the time when the ball is on the floor: t = –0.18 s and t = 1.7 s

•  The first answer tells when the ball would have been on the floor if it had always been on this trajectory; the second is the answer we want.

2 Significant Figures!


Summary

§  Position, velocity, and acceleration are the fundamental quantities describing motion. §  Velocity is the rate of change of position. §  Acceleration is the rate of change of velocity.

•  When acceleration is constant, simple equations relate position, velocity, acceleration, and time.

–  An important case is the acceleration due to gravity near Earth’s surface.

–  The magnitude of Earth’s gravitational acceleration is g = 9.8 m/s2 = 32 ft/s2.

�v(t) = at

�x(t) =1

2[v0 + v(t)]t

�x(t) = v0t+1

2at

2

v

2(t)� v

20 = 2a�x(t)


Break Time


§  Acceleration a (constant!)

§  Velocity v

§  Position x

v = v0 + at t =v � v0

a

Constant Acceleration and Graph Areas

x = x0 + v0t+1

2at

2

v =�x

�t

a =�v

�t�v = a�t

(area!!)

�x = area too!area = v0t

area = (1/2)(at)(t) = (1/2)at2

area = a�t


§  At the moment a traffic light turns green, a car that has been waiting at the intersection accelerates with constant acceleration of ac=3.2 m/s2. At that same moment, a truck whizzes by at a constant velocity of vt=20.0 m/s. §  How far beyond the traffic light does the car catch the truck? §  How fast is the car moving when it catches the truck?

§  Say the traffic light is x=0, so initial positions of both car and truck are xc0=xt0=0 m.

Using the Equations: Example 1 (5 minutes)

car

vc0 = 0 m/s

ac = 3.2 m/s2

xc = vc0t+1

2act

2 =1

2act

2

truck

vt0 = 20.0 m/s

at = 0.0 m/s2

xt = vt0t+1

2att

2 = vt0t

xc = xt when1

2act

2 = vt0t


§  At the moment a traffic light turns green, a car that has been waiting at the intersection accelerates with constant acceleration of ac=3.2 m/s2. At that same moment, a truck whizzes by at a constant velocity of vt=20.0 m/s. §  How far beyond the traffic light does the car catch the truck? §  How fast is the car moving when it catches the truck?

Using the Equations: Example 1 (cont)

xc = xt when1

2act

2 = vt0t

t =2vt0ac

=2(20.0 m/s)

3.2 m/s2= 12.5 s

vc = vc0 + act = act

vc = (3.2 m/s2)(12.5 s) = 40 m/s

x = vt0t = (20.0 m/s)(12.5 s) = 250 m


Using the Equations: Example 2

§  I toss a bean bag up in the air at v0=6.0 m/s. §  How far up from my hand does it go? §  How long does it take to come back to my hand?

•  (Assume my hand stays at the same height, x0=0.0 m).

§  We want to know x… but we don’t know t to “plug and chug”!

•  (This is typical of a frustration with the homework)

x = x0 + v0t+1

2at

2x0 = 0 m v0 = 6.0 m/s






§  You have another equation, and you know v=0 m/s at top.

§  If you have trouble going from the equation on the left to the equation on the right, please see me after class or email me

•  Calculus may not be a prerequisite for this class, but algebra is!


x = x0 + v0t+1

2at

2x0 = 0 m v0 = 6.0 m/s

v = v0 + at ) t =v � v0

a=

�v0a

(here)







§  You have another equation, and you know v=0 m/s at top.

§  Now substitute into the first equation…

x = x0 + v0t+1

2at

2x0 = 0 m v0 = 6.0 m/s

x� x0 = v0

✓�v0

a

◆+

1

2a

✓�v0

a

◆2

= �1

2

v

20

a

= x� x0

v = v0 + at ) t =v � v0

a=

�v0a

(here)




§  On the previous page we’d also figured out t in terms of v0


x� x0 = �1

2

v

20

a

x� x0 = �1

2

v

20

a

= �1

2

(6.0 m/s)2

(�9.8 m/s2)

�=

18.0 m2/s2

9.8 m/s2= 1.8 m

t =�v0a

=�6.0 m/s

(�9.8 m/s2)= 0.61 s




§  On the previous page we’d also figured out t in terms of v0

§  Be careful! The time back to my hand is TWICE this! (up/down)


x� x0 = �1

2

v

20

a

x� x0 = �1

2

v

20

a

= �1

2

(6.0 m/s)2

(�9.8 m/s2)

�=

18.0 m2/s2

9.8 m/s2= 1.8 m

t =�v0a

=�6.0 m/s

(�9.8 m/s2)= 0.61 s

t = time back to hand = 1.21 s


§  Acceleration a (constant!)

§  Velocity v

§  Position x

v = v0 + at t =v � v0

a

Huh, that looks a bit familiar…

x = x0 + v0t+1

2at

2

v =�x

�t

a =�v

�t�v = a�t

(area!!)

�x = area too!

t = 0.61 s

x� x0 = 1.8 m

v0 = 6.0 m/s


§  I want to toss a bean bag up in the air to just barely graze the ceiling that is 2.0m above my hand. §  How fast v0 do I have to throw the beanbag? §  If I toss the beanbag at t=0s and miss the catch on the way

back down, at what time does the beanbag hit the floor 1.5m below my hand?


§  We know v=0 at the ceiling, and a=g, and want to know v0…

but we don’t know t to “plug and chug”! •  (This is typical of a frustration with the homework again)

Using the Equations: Example 3

v = v0 + at ) v0 = v � at = �at (here)





§  In the previous problem we figured out an equation that eliminated time (only when vf=0 m/s):


v = v0 + at ) v0 = v � at = �at (here)

x� x0 = v0

✓�v0

a

◆+

1

2a

✓�v0

a

◆2

= �1

2

v

20

a

= x� x0

) v0 =p

�2a(x� x0)

v0 =p

�2(�9.8 m/s2)(2.0 m) = 6.3 m/s = v0





§  For the second part,

§  Uh oh… This is a quadratic equation. You can’t solve this using simple algebra – you have to remember the equation for the solutions of a quadratic.


x� x0 = �1.5 m

x� x0 = v0t+1

2at

2

(�1.5 m) = (6.3 m/s)t� (4.9 m/s2)t2


Quadratic Equation Refresher

§  For any quadratic equation of the form

where a,b,c are known values, the TWO values of x that “solve” this equation can be found with the equation Note that you should usually find that for physics problems otherwise the square root produces imaginary numbers (which don’t have good physical meanings!)

ax

2 + bx+ c = 0

x =�b±

⇥b

2 � 4ac

2a

b2 � 4ac > 0

This is general math: a is NOT acceleration x is NOT position






(�1.5 m) = (6.3 m/s)t� (4.9 m/s2)t2

(�4.9 m/s2)t2 + (6.3 m/s)t+ 1.5 m = 0

a = (�4.9 m/s2) b = (6.3 m/s) c = 1.5 m

t =�b±

⇥b2 � 4ac

2a=

�6.3 m/s± 8.3 m/s

�9.8 m/s2= 1.5 s,�0.20 s

Two solutions!


§  I drop a bean bag from my hand to the floor 1.5 m below my hand. §  How long does it take to reach the floor? §  What is the beanbag’s velocity at floor height?

•  (Assume floor height is x0=0.0 m).

Ponderable (10 minutes)

v = v0 + at x = x0 + v0t+1

2at

2


§  I drop a bean bag from my hand to the floor 1.5 m below my hand. §  How long does it take to reach the floor? §  What is the beanbag’s velocity at floor height?

•  (Assume floor height is x0=0.0 m).

§  Note that the equations don’t care about past or future!

Ponderable (10 minutes)

x = x0 + v0t+1

2at

2 ) �1.5 m =1

2at

2

x-x0=-1.5m v0=0 m/s

t =

s2(�1.5 m)

(�9.8 m/s2)= ±0.55 s

v = v0 + at v = (0 m/s) + (�9.8 m/s2)(±0.55 s) = ±5.4 m/s


§  I want to toss a bean bag up in the air to just barely graze the ceiling that is 2.0m above my hand, AND I’m throwing it to some unfortunate student who is 5.0m away from me. §  Here, catch! (*thud*) §  How is this different from just throwing the beanbag straight up

to graze the ceiling? •  Do I have to throw the one to the student with a higher or lower

velocity than the one straight up that grazes the ceiling? •  Which one stays in the air a longer time? •  At what angle from horizontal do I have to throw the one to the

student? More or less than 45 degrees?

Example 3 With a Twist (~5 minutes)


§  I want to toss a bean bag up in the air to just barely graze the ceiling that is 2.0m above my hand, AND I’m throwing it to some unfortunate student who is 5.0m away from me.

•  Do I have to throw the one to the student with a higher or lower velocity than the one straight up that grazes the ceiling?

•  Which one stays in the air a longer time? •  At what angle from horizontal do I have to throw the one to the

student? More or less than 45 degrees?

§  The time thing is tricky and nonintuitive. §  The vertical portion of the motion is the same for both cases

•  Therefore the time the bag is in the air is the SAME for both too! §  Two objects that have the same vertical motion take the same

time for that motion regardless of their horizontal motion! •  Example: a cannonball fired horizontally out of a cannon and

another cannonball dropped from that cannon take the same time to hit the ground.

Example 3 With a Twist (Explanation)

Higher!

Less, but how much less?


Onwards to Two Dimensional Motion

§  We’ll get further into examples of these equations next class §  Horizontal motion is easy: acceleration is zero! §  You’ve already done the hard part, vertical motion with gravity

Horizontal motion (no acceleration) Vertical motion (gravitational acceleration) Components of velocity, and vectors:

v

x

= v

x0

x = x0 + v

x0t

v

y

= v

y0 + at

y = y0 + v

y0t+1

2at

2

�v = vx

x+ vy

y

v =qv2x

+ v2y

x

y�v

vy

vx

University Physics 226N/231N Old Dominion University ...Aug 31, 2016 · Old Dominion University...

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Transcript of University Physics 226N/231N Old Dominion University ...Aug 31, 2016 · Old Dominion University...